hihoCoder 1122 二分图最大匹配 最大流

时间:2021-12-04 05:55:46

题意

  • 求二分图最大匹配

思路

  • 没有用匈牙利算法,而是试了试要带去的两个最大流模板。。
  • 就是加一个源点,一个汇点,源到二分图集合1的点加一个流量为1的边,集合1到集合2的边设置为流量1,集合2的点到汇点加入流量1的边

实现

  • 模板1实现
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <cstdio> 
#include <vector>
using namespace std;

const int MAXN = 2000;
const int maxn = 2000;
//最大流ISAP部分
const int MAXM = 20005;
const int INF = 0x3f3f3f3f;
struct Edge
{
    int to,next,cap,flow;
}edge[MAXM];
int tol;
int head[MAXN];
int gap[MAXN],dep[MAXN],pre[MAXN],cur[MAXN];
void init()
{
    tol = 0;
    memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int w,int rw = 0)
{
    edge[tol].to = v;
    edge[tol].cap = w;
    edge[tol].next = head[u];
    edge[tol].flow = 0;
    head[u] = tol++;
    edge[tol].to = u;
    edge[tol].cap = rw;
    edge[tol].next = head[v];
    edge[tol].flow = 0;
    head[v] = tol++;
}
int sap(int start,int end,int N)
{
    memset(gap,0,sizeof(gap));
    memset(dep,0,sizeof(dep));
    memcpy(cur,head,sizeof(head));
    int u = start;
    pre[u] = -1;
    gap[0] = N;
    int ans = 0;
    while(dep[start] < N)
    {
        if(u == end)
        {
            int Min = INF;
            for(int i = pre[u]; i != -1;i = pre[edge[i^1].to])
                if(Min > edge[i].cap - edge[i].flow)
                    Min = edge[i].cap - edge[i].flow;
            for(int i = pre[u];i != -1;i = pre[edge[i^1].to])
            {
                edge[i].flow += Min;
                edge[i^1].flow -= Min;
            }
            u = start;
            ans += Min;
            continue;
        }
        bool flag = false;
        int v;
        for(int i = cur[u];i != -1;i = edge[i].next)
        {
            v = edge[i].to;
            if(edge[i].cap - edge[i].flow && dep[v] + 1 == dep[u])
            {
                flag = true;
                cur[u] = pre[v] = i;
                break;
            }
        }
        if(flag)
        {
            u = v;
            continue;
        }
        int Min = N;
        for(int i = head[u];i != -1;i = edge[i].next)
            if(edge[i].cap - edge[i].flow && dep[edge[i].to] < Min)
            {
                Min = dep[edge[i].to];
                cur[u] = i;
            }
        gap[dep[u]]--;
        if(!gap[dep[u]])return ans;
        dep[u] = Min+1;
        gap[dep[u]]++;
        if(u != start)u = edge[pre[u]^1].to;
    }
    return ans;
}
//the end of 最大流部分

int vis[maxn];
vector<int> g[maxn];
#define pb push_back

void dfs(int u){
    for (int e=0;e<g[u].size();e++){
        int v = g[u][e];
        if (vis[v] == -1){
            vis[v] = (vis[u] ^ 1);
            dfs(v);
        }
    }
}

int main(){
    int n,m;
    cin>>n>>m;
    int u,v;
    for (int i=0;i<m;i++){
        scanf("%d%d",&u,&v);
        g[u].pb(v);
        g[v].pb(u);
    }
    memset(vis,-1,sizeof(vis));
    for (int i=1;i<=n;i++){
        if (vis[i] == -1){
            vis[i] = 0;
            dfs(i);
        }       
    }
    init();     
    for (int i=1;i<=n;i++){
        if (vis[i] == 0){
            addedge(0,i,1);
            for (int e=0;e<g[i].size();e++){
                v = g[i][e];
                addedge(i,v,1);
            }
        }
        else{
            addedge(i,n+1,1);
        }
    }
    cout << sap(0,n+1,n+2) << "\n";
    return 0;
}
  • 实现2
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <cstdio> 
#include <vector>
using namespace std;

const int maxn = 2000;
const int inf = 0x3f3f3f3f;
struct EdmondsKarp{
    //用到外部变量maxn,maxm,具体使用不非得按照模板 
    int n,m;
    int g[maxn][maxn];
    int flow[maxn];
    int path[maxn];
    queue<int> q; 
    void init(int nn=0)
    {
        n = nn;
        memset(g,0,sizeof(g));
        memset(flow,0,sizeof(flow));
        memset(path,-1,sizeof(path));
    }
    void addOne(int u,int v,int yuan)
    {
        g[u][v] += yuan;
        m += 2; 
    }

    int bfs(int s,int t)
    {
        int i;
        memset(path,-1,sizeof(path));
        flow[s] = inf;
        //如果不加这句话,bfs内部就要一共三个判断 
        path[s] = s; 
        q.push(s);
        while(q.size()>0)
        {
            int tmp = q.front();
            q.pop();
            //这里i从0开始还是1开始,视情况而定 
            for(i=0;i<n;i++)
            {
                if(g[tmp][i]>0 && path[i]==-1)
                {
                    flow[i] = min(g[tmp][i],flow[tmp]);
                    path[i] = tmp;
                    q.push(i);
                    if(i==t)
                    {
                        break;
                    }
                }   
            }
            if(path[t] != -1)
                break;
        }
        while(q.size()>0)
            q.pop();
        if(path[t] == -1)
            return 0;
        else    
            return flow[t];
    }

    int go(int s,int t)
    {
        int ret = 0;
        while(bfs(s,t))
        {
            int now = path[t];
            int pre = t;
            ret += flow[t];
            while(pre != s)
            {
                g[now][pre] -= flow[t];
                g[pre][now] += flow[t];
                pre = now;
                now = path[now];
            }
        }
        return ret;
    }

}edk;


int vis[maxn];
vector<int> g[maxn];
#define pb push_back

void dfs(int u){
    for (int e=0;e<g[u].size();e++){
        int v = g[u][e];
        if (vis[v] == -1){
            vis[v] = (vis[u] ^ 1);
        }
    }
}

int main(){
    int n,m;
    cin>>n>>m;
    int u,v;
    for (int i=0;i<m;i++){
        scanf("%d%d",&u,&v);
        g[u].pb(v);
        g[v].pb(u);
    }
    memset(vis,-1,sizeof(vis));
    for (int i=1;i<=n;i++){
        if (vis[i] == -1){
            vis[i] = 0;
            dfs(i);
        }       
    }
    edk.init(n+2);      
    for (int i=1;i<=n;i++){
        if (vis[i] == 0){
            edk.addOne(0,i,1);
            for (int e=0;e<g[i].size();e++){
                v = g[i][e];
                edk.addOne(i,v,1);
            }
        }
        else{
            edk.addOne(i,n+1,1);
        }
    }
    cout << edk.go(0,n+1) << "\n";
    return 0;
}